Help for new user designing off grid power system

Beto, welcome to the forum.

I have moved your question to its own thread--Will be easier and less confusing to have the discussion focused on just your needs.

In general, I/we recommend that you define your loads first, and describe what they are. Then design a system that will (hopefully) meet your loads/power requirements. Otherwise, you can end up with a system that does not really meet your needs and/or costs more that it should.

For you--Do you already have all of the hardware? Anyway--assuming your system is somewhere in the the US southwest, and your loads are 400 Watts for 5 hours per day...

First the battery bank. For lead acid batteries 1-3 days of storage and 50% maximum discharge (for longer battery life) seems to be a "useful" range. And 2 days/50% maximum discharge usually seems to be optimum:

• 400 Watts * 5 hours * 1/12 volt battery * 1/0.85 AC inverter eff * 2 days storage * 1/0.50 maximum discharge = 784 AH @ 12 volt battery bank

That is about the maximum I would suggest for a 12 volt battery bank. For various reasons, you might find a 24 volt system is a bit easier system to build/maintain (>800 AH battery banks need very heavy copper cable for high currents, multiple solar charge controllers, and larger AH cells/more parallel batteries).

If you have 220 AH @ 6 volt batteries, you would need:

• 784 AH / 225 AH per string = 3.5 or 4 parallel string * 2x 6 volt batteries per string

So, you have a bit more battery capacity than I wohttp://forum.solar-electric.com/forum/solar-electric-power-wind-power-balance-of-system/off-grid-solar-battery-systems/342809-help-for-new-user-designing-off-grid-power-systemuld suggest as optimal... And placing lots of Lead Acid batteries in parallel can make for some maintenance headaches (parallel strings of batteries do not always share current well, you should have a fuse/circuit breaker per battery, more cells to check water level/specific gravity in, more cables, etc.).

But since you already have the batteries, we recommend 5% to 13% rate of charge for solar. To properly charge such a battery bank:

• 1,125 AH * 14.5 volts charging * 1/0.77 panel+controller deratings * 0.05 rate of charge = 1,059 Watt array minimum (weekend/seasonal usage)
• 1,125 AH * 14.5 volts charging * 1/0.77 panel+controller deratings * 0.10 rate of charge = 2,119 Watt array nominal (full time off grid)
• 1,125 AH * 14.5 volts charging * 1/0.77 panel+controller deratings * 0.13 rate of charge = 2,754 Watt array "cost effective" maximum

And based on the amount of sun you receive and the loads you have--Using PV Watts, for Dagget California (need your rough location, just taking a guess for the calculations), fixed array facing south, tilted 35 degrees from horizontal:

Month    Solar Radiation (kWh/m 2/day)
1      5.59     
2      6.03     
3      7.10     
4      7.74     
5      7.42     
6      7.42     
7      7.33     
8      7.40     
9      7.34     
10      6.76     
11      5.78     
12      5.20     
Year      6.76

Using 5.2 hours of sun minimum (lots of sun in this desert):

• 400 Watts * 5 hours per day load * 1/0.52 typical off grid system eff * 1/5.2 hours of sun for December = 740 Watt array minimum

And you have "11, 75w ,12v panels & 3, 120w, 12v panels & 3 , 15w, 12v panels". Don't even bother with the 15 watt panels, assuming the rest of the panels have Vmp~17.5 or so voltage, and you put them in parallel = 1,110 Watt array.

At this point, put all of the panels in parallel (you should have a combiner box with fuse/breaker per solar panel string to reduce the chance of fire if one panel gets shorted and the other feed current into the shorted cable).

You have the minimum amount of current recommended (>5% rate of charge) for your relatively large battery bank. And, if I got your location correct, because you have lots of sun in the southwestern desert--You can probably get away with a somewhat smaller solar array (5%) than I would suggest for a full time off grid system (10% or more is a typical recommendation).

The AC inverter may be a bit large for your needs (1,500 Watt for a 400 watt load)--Large inverters can have fairly high self consumption (10-20 watts+) just being turned on (tare losses). If you run the inverter 24 hours per day and it has 20 watts self consumption:

• 20 Watts * 24 hours per day = 480 Watt*Hours per day of self consumption

If you use 2,000 Watt*Hours per day, then almost 20% of your system load would be powering the AC inverter. If you can turn off the inverter and/or if it has "search mode", you could reduce the inverters "load" on your battery bank.

Anyway--My first guess at your system configuration/design.

-Bill